Question

If one root of the equation $4x^2-6x+p=0$ is $q+2i$, where $p,q\in R$ then $p+q=$

• A. $10$
• B. $19$
• C. $-24$
• D. $-32$

Answer 1

The correct option is B

$19$

Explanation for the correct option.

Step 1: Information required for the solution

We know that the quadratic equation of the form $x^2+ax+b=0$ has its sum of roots equals to the coefficient of $x$ that is $a$ and product of roots equals to the constant $b$.

For this, we need to convert the given equation in the form $x^2+ax+b=0$. For that divide both sides of the equation $4x^2-6x+p=0$ by $4$.

The equation becomes $x^2-\frac{3}{2}x+\frac{p}{4}=0$.

Now, let the roots of this equation be $\alpha \text{and}\beta$, then

$\alpha +\beta =\frac{3}{2}\dots \left(1\right)$

$\alpha \beta =\frac{p}{4}\dots \left(2\right)$

Step 2: Calculation for the value of $p\text{and}q$

As one root of the equation is $q+2i$ then let this be equal to $\alpha$ and the other root $\beta$ will be equal to $q-2i$.

$\begin{array}{rcl}\therefore \alpha +\beta & =& q+2i+q-2i\\ & =& 2q\end{array}$

From equation $\left(1\right)$, we have $\alpha +\beta =\frac{3}{2}$

$\begin{array}{rcl}\therefore 2q& =& \frac{3}{2}\\ & \Rightarrow & q=\frac{3}{4}\end{array}$

Now, we need to determine the value of $p$,

$\begin{array}{rcl}\therefore \alpha \beta & =& \left(q+2i\right)\left(q-2i\right)\\ & =& q^2-\left(4\right)\left(-1\right)\\ & =& q^2+4\end{array}$

Form equation $\left(2\right)$, we have $\alpha \beta =\frac{p}{4}$

$\begin{array}{rcl}\therefore \frac{p}{4}& =& q^2+4\\ & =& {\left(\frac{3}{4}\right)}^2+4\\ & =& \frac{9}{16}+4\\ & =& \frac{73}{16}\end{array}$

Therefore, the value of $p$ is $\frac{73}{4}$.

Step 3: Calculation of the value of $p+q$

Finally, the value of $p+q$ is

$\begin{array}{rcl}\therefore p+q& =& \frac{73}{4}+\frac{3}{4}\\ & =& \frac{76}{4}\\ & =& 19\end{array}$

Hence, the correct option is (B).